Question

In a $L-R$ circuit, the value of $L$ is $(0.4 / \pi) H$ sar the value of $R$ is $30 \,\Omega$. If in circuit, an alternating emf of $200 \,V$ at $50\, cycle / s$ is connected, the impedance of the circuit and current will be

• A. $11. \, \Omega , 17.5\,A$
• B. $30. \, \Omega , 6.5\,A$
• C. $40. \, \Omega , 5 \,A$
• D. $50 \, \Omega , 4\, A$

Answer 1

Answer: (D)

Impedance of $L-R$ circuit is
$Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{R^{2}+(\omega L)^{2}}$
$=\sqrt{R^{2}+(2 \pi f L)^{2}}$
Given, $L=\frac{0.4}{\pi} H, R=30\, \Omega, f=50$ cycle/s
So, $Z=\sqrt{(30)^{2}+\left(2 \pi \times 50 \times \frac{0.4}{\pi}\right)^{2}}$
$=\sqrt{(30)^{2}+(40)^{2}}=\sqrt{900+1600}=50\, \Omega$
and current $I=\frac{V}{Z}=\frac{200}{50}=4 \,A$